Question

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of

Answer 1

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = $\frac{mass}{molar mass}$  

Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = $\frac{18.1}{143.2}$   = 0.13moles

Number of moles of NH₃   = $\frac{90.4}{17}$   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

    Mass  = 0.13 x (2 x 14) = 3.64g